Question

The family of circles x2 + y2 + ax + ay = 2 + 2a for different values of 'a' pass
(a) One fixed point
(b) Two fixed point
(c) More than two fixed point
(d) None of these​

Answer 1

Answer:

non of these

Step-by-step explanation

Answer 2

The family of circles x² + y² + ax + ay = 2 + 2a for different values of 'a' pass will pass through two fixed points   (-√2 , - √2) & (√2 , √2)

Step-by-step explanation:

x² + y²  + ax + ay = 2 + 2a

=> (x + a/2)²  - a²/4  + (y + a/2)² - a²/4    = 2 + 2a

=> (x + a/2)² + (y + a/2)²  = a²/4  + 2a + 2

=>  (x + a/2)² + (y + a/2)²  = (a/√2   + 2)²

Center = (-a/2 , -a/2)   Radius  =  (a/√2   + 2)

if a = 0

then Circle center at origin will  radius  = 2

hence circle has point  (-√2 , - √2) , (-√2 , √2) , (√2 , -√2) & (√2 , √2)

For a = + ve =>  cenetr lies in 3rd Quadrant

Distance of  (√2 , √2)  from center  =  √((√2 + a/2)² +  (√2 + a/2)²)

= (2 + a√2) = Radius

For a = -ve  =>  cenetr lies in 1st Quadrant

Distance of  (-√2 , -√2)  from center  = √((-√2 - a/2)² +  (√2 - a/2)²)

= (2 + a√2) = Radius

Hence The family of circles x² + y² + ax + ay = 2 + 2a for different values of 'a' pass will pass through two fixed points   (-√2 , - √2) & (√2 , √2)

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